[Poj 2488] A Knight's Journey

A Knight’s Journey
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 38352 Accepted: 13017

Description
Background
The knight is getting bored of seeing the same black and white squares again and again and has decided to make a journey
around the world. Whenever a knight moves, it is two squares in one direction and one square perpendicular to this. The world of a knight is the chessboard he is living on. Our knight lives on a chessboard that has a smaller area than a regular 8 * 8 board, but it is still rectangular. Can you help this adventurous knight to make travel plans?

Problem
Find a path such that the knight visits every square once. The knight can start and end on any square of the board.

Input
The input begins with a positive integer n in the first line. The following lines contain n test cases. Each test case consists of a single line with two positive integers p and q, such that 1 <= p * q <= 26. This represents a p * q chessboard, where p describes how many different square numbers 1, … , p exist, q describes how many different square letters exist. These are the first q letters of the Latin alphabet: A, …

Output
The output for every scenario begins with a line containing “Scenario #i:”, where i is the number of the scenario starting at 1. Then print a single line containing the lexicographically first path that visits all squares of the chessboard with knight moves followed by an empty line. The path should be given on a single line by concatenating the names of the visited squares. Each square name consists of a capital letter followed by a number.
If no such path exist, you should output impossible on a single line.

Sample Input

3
1 1
2 3
4 3

Sample Output

Scenario #1:
A1

Scenario #2:
impossible

Scenario #3:
A1B3C1A2B4C2A3B1C3A4B2C4

#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int vis[27][27];
int pre[27][27][2];
int n,m;
int T;
int ans[27][2];
int flag;
int di[8][2]={{-2,-1},{-2,1},{-1,-2},{-1,2},{1,-2},{1,2},{2,-1},{2,1}};
bool can(int x,int y)
{
    if(vis[x][y]) return 0;
    if(x<=0||y<=0||x>m||y>n) return 0;
    return 1;
}
void outt(int x,int y)
{
    //cout<<x<<"   "<<y<<endl;
    int cnt=0;  
    while(x!=0)
    {
        cnt++;
        ans[cnt][0]=x;
        ans[cnt][1]=y;

        int xx=pre[x][y][0];
        y=pre[x][y][1];
        x=xx;
        //cout<<x<<"            "<<y<<endl;
    }
    for(int i=n*m;i>=1;i--)
    {
        cout<<char(ans[i][0]-1+'A')<<ans[i][1];
    }
    cout<<endl;


}
void dfs(int x,int y,int cnt)
{
    vis[x][y]=1;

    cnt--;
    if(flag) return;
    if(cnt==0)  
    {
        flag=1;
        outt(x,y);
        return ;
    }
    int mx,my;
    for(int i=0;i<8;i++)
    {
        int mx=di[i][0]+x;
        int my=di[i][1]+y;
        if(can(mx,my))
        {   
            pre[mx][my][0]=x;
            pre[mx][my][1]=y;
                dfs(mx,my,cnt);
        }

    }
    vis[x][y]=0;
    return ;
}
int main()
{
    scanf("%d",&T);
    for(int jj=1;jj<=T;jj++)
    {
        scanf("%d%d",&n,&m);
        printf("Scenario #%d:\n",jj);
        memset(vis,0,sizeof(vis));
        flag=0;
        int cnt=n*m;

            dfs(1,1,cnt);

        if(flag==0) printf("impossible\n");
        printf("\n");
    }       
}